poj1273 网络流入门题 dinic算法解决，可作模板使用

Drainage Ditches

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 62078		Accepted: 23845

Description

Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches so that Bessie's clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch.
Farmer John knows not only how many gallons of water each ditch can
transport per minute but also the exact layout of the ditches, which
feed out of the pond and into each other and stream in a potentially
complex network.

Given all this information, determine the maximum rate at which
water can be transported out of the pond and into the stream. For any
given ditch, water flows in only one direction, but there might be a way
that water can flow in a circle.

Input

The input includes several cases.
For each case, the first line contains two space-separated integers, N
(0 <= N <= 200) and M (2 <= M <= 200). N is the number of
ditches that Farmer John has dug. M is the number of intersections
points for those ditches. Intersection 1 is the pond. Intersection point
M is the stream. Each of the following N lines contains three integers,
Si, Ei, and Ci. Si and Ei (1 <= Si, Ei <= M) designate the
intersections between which this ditch flows. Water will flow through
this ditch from Si to Ei. Ci (0 <= Ci <= 10,000,000) is the
maximum rate at which water will flow through the ditch.
Output

For each case, output a single integer, the maximum rate at which water may emptied from the pond.
Sample Input

5 4
1 2 40
1 4 20
2 4 20
2 3 30
3 4 10

Sample Output

50

Source

USACO 93

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<queue>
#include<algorithm>
using namespace std;
int edge[300][300];//邻接矩阵
int dis[300];//距源点距离,分层图
int start,end;
int m,n;//N:点数;M,边数
int bfs(){
memset(dis,-1,sizeof(dis));//以-1填充
dis[1]=0;
queue<int>q;
q.push(start);
while(!q.empty()){
int u=q.front();
q.pop();
for(int i=1;i<=n;i++){
if(dis[i]<0&&edge[u][i]){
dis[i]=dis[u]+1;
q.push(i);

}
}
}
if(dis
>0)
return 1;
else
return 0;//汇点的DIS小于零,表明BFS不到汇点
}
//Find代表一次增广,函数返回本次增广的流量,返回0表示无法增广
int find(int x,int low){//Low是源点到现在最窄的(剩余流量最小)的边的剩余流量
int a=0;
if(x==n)
return low;//是汇点
for(int i=1;i<=n;i++){
if(edge[x][i]>0&&dis[i]==dis[x]+1&&//联通,,是分层图的下一层
(a=find(i,min(low,edge[x][i])))){//能到汇点(a <> 0)
edge[x][i]-=a;
edge[i][x]+=a;
return a;
}

}
return 0;
}
int main(){
while(scanf("%d%d",&m,&n)!=EOF){
memset(edge,0,sizeof(edge));
for(int i=1;i<=m;i++){
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
edge[u][v]+=w;
}
start=1;
end=n;
int ans=0;
while(bfs()){//要不停地建立分层图,如果BFS不到汇点才结束
ans+=find(1,0x7fffffff);//一次BFS要不停地找增广路,直到找不到为止
}
printf("%d\n",ans);
}
return 0;
}